EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
10th Edition
ISBN: 8220106906149
Author: Jewett
Publisher: Cengage Learning US
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Chapter 17, Problem 50CP

(a)

To determine

To show: The left hand side of the given equation is 0 if m is even and 4Amω if m is odd.

(a)

Expert Solution
Check Mark

Answer to Problem 50CP

The left hand side of the given equation is 0 if m is even and 4Amω if m is odd.

Explanation of Solution

Given info: The frequency of wave is f , the amplitude of wave is A . The period of wave is described by y(t)={A0<t<T2AT2<t<T . The given equation is y(t)=n(Ansinnωt+Bncosnωt) .

Write the expression of the y(t) .

y(t)=n(Ansinnωt+Bncosnωt)

Multiply sinmωt both sides in above equation.

y(t)sinmωt=(Ansinnωt+Bncosnωt)(sinmωt)

Integrate the above equation over on period of time T .

0Ty(t)sinmωt=[n0TAn(sinnωt)(sinmωt)dt+0TBn(cosnωT)(sinnωT)dt] (1)

The term y(t) is a positive constant A for half of the period T and A for next half period.

The left side of the equation is,

0Ty(t)sinmωt=0T2Asinmωt+0T2Asinmωt=Amω[cosmω(T2)cos0]+Amω[cosmωTcosmω(T2)]={2Amωmodd0meven}+{2Amωmodd0meven}={4Amωmodd0meven}

Conclusion:

Therefore, the left hand side of the given equation is 0 if m is even and 4Amω if m is odd.

(b)

To determine

To show: The terms of the right hand side of the given equation involving Bn are equal to zero.

(b)

Expert Solution
Check Mark

Answer to Problem 50CP

The terms of the right hand side of the given equation involving Bn are equal to zero.

Explanation of Solution

Given info: The frequency of wave is f , the amplitude of wave is A . The period of wave is described by y(t)={A0<t<T2AT2<t<T . The given equation is y(t)=n(Ansinnωt+Bncosnωt) .

The part of the right hand of the equation (1) involving Bn is,

0TBn(cosnωT)(sinnωT)dt=12Bn0T[sin(nωt+mωt)sin(nωtmωt)]dt=0+0=0

Conclusion:

Therefore, the terms of the right hand side of the given equation involving Bn are equal to zero.

(c)

To determine

To show: The terms of the right hand side of the given equation involving An are equal to zero.

(c)

Expert Solution
Check Mark

Answer to Problem 50CP

The terms of the right hand side of the given equation involving An are equal to zero.

Explanation of Solution

Given info: The frequency of wave is f , the amplitude of wave is A . The period of wave is described by y(t)={A0<t<T2AT2<t<T . The given equation is y(t)=n(Ansinnωt+Bncosnωt) .

The part of the right hand of the equation (1) involving An is,

0TBn(cosnωT)(sinnωT)dt=12An0T[cos(nωtmωt)cos(nωt+mωt)]dt=12An0T[cos(nm)ωtcos(n+m)ωt]dt=12An[1(nm)ωsin(nm)ωt1(n+m)ωsin(n+m)ωt]0T=0

The integration of the sinθ over the period of cycle is zero.

Conclusion:

Therefore, the terms of the right hand side of the given equation involving An are equal to zero.

(d)

To determine

To show: The entire right hand side of the equation reduces to 12AmT .

(d)

Expert Solution
Check Mark

Answer to Problem 50CP

The entire right hand side of the equation reduces to 12AmT .

Explanation of Solution

Given info: The frequency of wave is f , the amplitude of wave is A . The period of wave is described by y(t)={A0<t<T2AT2<t<T . The given equation is y(t)=n(Ansinnωt+Bncosnωt) .

The right hand side of the equation (1) is,

[0TAn(sinnωt)(sinmωt)dt+0TBn(cosnωT)(sinnωT)dt]=[0TAn(sinnωt)(sinmωt)dt+120TBn(sin2nωT)dt]=0TAn(sinnωt)(sinmωt)dt+0

The integer n=m .

Substitute m for n in above equation.

0TAn(sinnωt)(sinmωt)dt=0TAm(sinmωt)(sinmωt)dt=12Am0T[cos(mm)ωT]dt=12Am0T1dt=12AmT

Conclusion:

Therefore, the entire right hand side of the equation reduces to 12AmT .

(e)

To determine

To show: The Fourier series expansion for a square wave is y(t)=n4Anπsinnωt .

(e)

Expert Solution
Check Mark

Answer to Problem 50CP

The Fourier series expansion for a square wave is y(t)=n4Anπsinnωt .

Explanation of Solution

Given info: The frequency of wave is f , the amplitude of wave is A . The period of wave is described by y(t)={A0<t<T2AT2<t<T . The given equation is y(t)=n(Ansinnωt+Bncosnωt) .

The general equation of y(t) is,

y(t)=Ansinnωt (2)

From the part (a), The value of 0Ty(t)sinmωtdt=4Amω when m is odd.

From the part (d), the value of 0Ty(t)sinmωtdt=12AmT , when n=m .

Therefore,

4Anω=12AmTAm=8AnωT

Substitute 2πT for ω in above equation.

Am=8An(2πT)T=4Anπ

Substitute Am for An in equation (2).

y(t)=Amsinnωt

Substitute 4Anπ for Am in above equation.

y(t)=4Anπsinnωt

Conclusion:

Therefore, the Fourier series expansion for a square wave is y(t)=n4Anπsinnωt .

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Chapter 17 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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